CS Basics 3) Encoding Numbers

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1 CS Basics 3) Encoding Numbers E.Benoist, C.Fuhrer, C.Grotho, L. Ith, P.Mainini Fall Term Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 1

2 Contents Binary vs. Text Files Endianness Negative Numbers Introduction to Floating Point Fixed Point Numbers Floating Point Numbers, IEEE 754 Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 2

3 Binary vs. Text Files Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 3

4 Binary vs. Text Files Dierent le formats (examples) Images: jpeg, gif, png, svg,... Documents: pdf, doc, ppt, xls,... Executables (Programs): exe, dll, so, class,... Plain text: txt, html, tex,... Source code: asm, c, cpp, java,... Two primary types Binary les (images, documents, executables,... ) Text les (plain text, source code,... ) Distinction not always strict svg is a format for vector graphics based on XML and can thus be considered a text le Oce documents are often also a form of compressed XML... Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 4

5 Text Files Dierent character encodings 7 bit / US-ASCII ISO-8859-* UTF-8... How to work with text les? Using any text editor or IDE vim, emacs, Notepad (windows), gedit, kate, sublime, atom,... Eclipse, NetBeans, IntelliJ, kdevelop,... Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 5

6 Binary Files I A lot of dierent le formats Open and proprietary Encoding and structure depend on type! Executable les (Programs) Windows: exe GNU/Linux: elf (32 bits), elf64 (64-bit) Contain machine instructions and data Can be analyzed with a decompiler or disassembler Images jpeg: format family, lossy compression gif: 8-bit color, animations possible png: 32-bit color, no animations, modern lossless compression (good for screenshots) svg: vector graphics, not a binary le type (see Slide 4) Can be manipulated with libraries and image editors Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 6

7 Binary Files II How to analyze any type of binary data? Using a hex editor Editors: Bless, xxd,... Bless functionality Display binary and textual representation at the same time Various conversions to dierent representations Can read and edit any le also executable les Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 7

8 Binary Files III Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 8

9 Binary Files IV Data is ultimately always encoded in binary Example Capital letter S is encoded with 0x53 (ASCII/UTF-8) For the computer, this is a set of 8 bits: Can also be part of a dierent number 0x53 may be interpreted as decimal 83 0x53_61 may be interpreted as the decimal 21'345 0x53_61_6D_0A may be interpreted as the decimal 1'398'893'834 0x53_61_6D_0A_77_61_73_0A may be interpreted as the oating point number But: This pattern can also mean anything else Can be part of an instruction Can be any data (object, address,... )... Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 9

10 Endianness Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 10

11 Endianness Having multiple bytes of data... Example: 0x53 0x61 0x6D 0x0A 0x77 0x61 0x73 0x0A Can be read/interpreted left-to-right (like European languages) 0x53 0x61 0x6D 0x0A 0x77 0x61 0x73 0x0A but can also be read/interpreted right-to-left (like in Arabic) 0x0A 0x73 0x61 0x77 0x0A 0x6D 0x61 0x The same data but dierent meaning! Which number does the following data represent? 0x53616D0A A? 0x0A A6D6153? Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 11

12 Big Endian vs. Little Endian Big Endian Data is stored with the most signicant byte (MSB) rst Example: Number 0x20A1 is stored as follows 0x20 0xA1 0 1 Little Endian Data is stored with the least signicant byte (LSB) rst Example: Number 0x20A1 is stored as follows 0xA1 0x Endianness is dependent of CPU architecture Intel x86 uses little endian Other architectures may use big endian Some CPUs even support both Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 12

13 There are big differences at stake here! The two bytes that begin our example Bern University of Applied text file Sciences represent Berner the decimal Fachhochschule number Haute 21,345école in aspécialisée big endian bernoise system, but 24, Big Endian vs. Little Endian Big Endian Offset Increases Little Endian Offset Increases Bytes in storage Most Least Significant Significant Byte Byte 16-bit hexadecimal 53 Most Least Significant Significant Byte Byte Unsigned decimal equivalent Figure 5-5: Big endian vs. little endian for a 16-bit value

14 Negative Numbers Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 14

15 How to Represent Negative Numbers? So far, we have seen positive numbers Example: 15 In memory: 0x0F (8 bit) or 0x F (32 bit) What about negative numbers? Example: -15 In memory:??? Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 15

16 Idea: Sign Bit What if we use one bit for the sign? Taking the rst bit as sign bit is still 0x000F (16 bit) -15 would be 0x800F Problems Two representations of 0 exist: 0x0000 and 0x8000 Standard (bit-level) addition does not work: 15 + (-15) = 0x000F + 0x800F = 0x801E = -30 Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 16

17 One's Complement One's Complement The negative number is obtained by inverting all the bits 15 = 0x000F 15 = B -15 is represented by B -15 = 0xFFF0 Problems We still have two representations of 0: 0x0000 and 0xFFFF Addition still does not work: 15 = 0x000F and -5 = 0xFFFA. The sum is 0x10009 = 9, if we ignore the overow Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 17

18 Two's Complement Negative numbers must respect arithmetic 15 + ( 5) = 10 Idea: X is represented by 2 n X, with n being the bit-width X + ( X ) = X + (2 n X ) = 0 with overow Example 1 Representation of 1 in 16 bits = is written 0xFFFF 1 + ( 1) = 0x10000 (overow!) Example 2 Representation of 20 in 16 bits = = hard to compute Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 18

19 Two's Complement Method for easy computation of two's complement Take the binary representation of X Invert all the bits of X Add 1 Example 2 again Want: = B... inverting all the bits B... adding B = 0xFFEC Arithmetic works! 15 + ( 5) = B B = B Ignoring overow: B = 10 Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 19

20 Introduction to Floating Point Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 20

21 Floating Point Arithmetic In the early days, CPUs where only able to handle integer arithmetic. However, for scientic and also everyday purposes, real numbers (R) are required. The data types for working with real numbers have special properties that every programmer must know. Representation of real numbers and oating point arithmetic is standardized in the IEEE 754 standard, which is supported by most programming languages and CPUs nowadays. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 21

22 Fixed Point Numbers Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 22

23 Fixed Point Numbers How to represent the decimal number in xed point arithmetic? Remember: Every number in base b can be written as x = i= d i b i where d i are the respective digits of the number and b is the base. This way, the number may be written as the sum: = Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 23

24 Binary Fixed Point Numbers As the number exists independently of any base, it must be possible to obtain a binary representation. We proceed as follows: 1. First, the integral part of the number is converted to binary 543 = B 2. Then, the fractional part of the number is converted as well = 0.001B The whole, xed point binary number is then = B Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 24

25 Converting to Fixed Point Binary I First, we convert the integer part as usual, using division by 2 with remainder: Example: = B Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 25

26 Converting to Fixed Point Binary II To convert the fractional (or decimal) part, we do the opposite and multiply by two until we have no remainder: Example: = 0.001B Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 26

27 Some Mathematical Considerations A rational number (Q) is either nite or innite periodic. The property of rationality is independent of the base used to represent a number. A rational number may have a nite representation in one base and a periodic innite representation in another base. For example, the number 1 has an innite periodic represention in 3 base 10, but may be written as in base 3! Irrational numbers (R \ Q) are innite and non-periodic. They do not have a nite representation in any base. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 27

28 Innite Periodic Number Example 0.1 is nite in decimal but not in binary: > > > > = B Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 28

29 Floating Point Numbers, IEEE 754 Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 29

30 Scientic Notation The main drawback of xed point numbers is that their size depends on their magnitude and their precision. In engineering, most of the time, the precision can be reduced as the magnitude increases. This is the concept of signicant digits. On pocket calculators, this is usually known as scientic notation. Examples: (= 13440) (= ) (= 4.430) All theses numbers have 4 signicant digits (called the mantissa). The exponent (10 x ) scales the numbers. With 4 signicant digits, the mantissa is always between and Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 30

31 The IEEE 754 Standard The IEEE 754 standard used for binary representation of oating point numbers is based on the same idea: The mantissa is a number 1.0 and < 2.0. The exponent is now an exponent of 2. It may take positive or negative values. Maximum exponent and precision (number of signicant digits in the mantissa) depends on the encoding format, see Slide 38. The standard also introduces special values (negative innity, positive innity, not a number, zero). It also contains rules for rounding. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 31

32 IEEE 754, An Example Example: Encoding the decimal number 0.5 (= ) in IEEE 754: = The rst bit is the sign bit. Its value is 0 for positive numbers and 1 for negative numbers. The exponent, in this case 2 1, is encoded with a so-called bias. The mantissa is 1 (not encoded due to the hidden bit ). Java). 1 The number is encoded in single precision format (e.g. float in C or Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 32

33 IEEE 754, The Exponent The exponent can be positive or negative. It is encoded as a signed integer with bias, i.e. without using two's complement! The bias is added to the original exponent and corresponds to half the magnitude of the exponent minus one. For example, if the exponent is 8 bits, its magnitude is 2 8 = 256 and the bias is 2 (8 1) 1 = 127. Two exponents, and FF 16, are reserved to increase precision for very small numbers and to represent special values. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 33

34 IEEE 754, The Mantissa Contains a value in the interval 1 mantissa < 2. As the mantissa is always 1, the bit for the 0th power may be ommitted, increasing precision. This is called the hidden bit. Due to this, the mantissa is 0 in the given example. Exception: For denormalized numbers (exponent 0), the mantissa m is 0 < mantissa < 1! This is used for very small numbers (below ). The bits of the mantissa represent the sum of negative powers of 2. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 34

35 IEEE 754, Values The following table lists possible values for IEEE 754 encoded numbers: Exponent Mantissa Value Description = 0 0 Zero > 0 ±0.m 2 (1 b) Denormalized to FE 16 any ±1.m 2 (e b) Normalized FF 16 = 0 ± ± Innity FF 16 > 0 NaN Not a Number (NaN) Remarks: e is the exponent, b the bias value. There are two possible representations for the value 0 (+0 and -0). Exponents and FF 16 represent special values! All other exponents (01 16 to FE 16 ) are normalized values. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 35

36 Comments on Special Values The value zero needs a special representation as it is not possible to represent it using the standard encoding (see example above). Innity represents numbers whose magnitude cannot be encoded. However, arithmetical operations where one operand is innity are well dened by IEEE 754 (see next slide). Operations where one operand has the value NaN cause an error. Warning: Handling of denormalized values is hardware and implementation dependent and may lead to performance issues. This must be considered when using them. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 36

37 IEEE 754 Operations with Special Values Operation Result x/ ± 0 ± ± ± ± non zero / 0 ± ±0/ ± 0 NaN + NaN ± / ± NaN ± 0 NaN Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 37

38 IEEE 754 Formats IEEE 754 (2019) denes three basic formats for binary oating point numbers: 2 Single precision, double precision and quadruple precision. They dier only by the number of bits required to store them: Sign Exponent Mantissa Bias Single precision (32 bit) Double precision (64 bit) Quadruple precision (128 bit) Note: Due to the hidden bit, precision equals to mantissa Formats with dierent bit widths, as well as decimal formats (using radix 10) are also specied as non-basic formats and are not treated here. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 38

39 IEEE 754 Rounding IEEE 754 has ve dierent rounding modes. The rst two round to nearest, the others are directed roundings. Round to nearest, ties to even Round to nearest encodable value. If exactly midway, round to the nearest value where the least signicant bit of the mantissa is 0. Round to nearest, ties away from zero Round to nearest also, but round up (positive numbers) or down (negative numbers) Round toward 0 Rounds to 0 (truncation) Round toward + Always round up. Round toward Always round down. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 39

40 Examples for Rounding Rounding Mode to nearest, ties to even to nearest, ties away from zero toward toward toward Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 40

41 Comparing Floating Points Values Many oating point operations are not associative due to rounding! That is, the same expression, but computed in a dierent order, may create dierent results! This makes the comparison of oating point values particularly tricky and complicated. Moreover, dierent compilers, even though implementing the same standard, may yield dierent results! Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 41

42 Problem: Absorbtion (C) Absorption occurs when working with numbers with large dierences in magnitude: #include <stdio.h> int main(void) { float val1 = 100.0, val2 = 0.05, val3 = 0.05; } printf("sum 1: %f\n", (val1 + val2) + val3); printf("sum 2: %f\n", val1 + (val2 + val3)); >./absorbtion sum 1 : sum 2 : Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 42

43 Problem: Absorbtion (Java) public class Absorbtion { public static void main (String args[]) { float val1 = 100.0f; float val2 = 0.05f; float val3 = 0.05f; System.out.println("sum 1 : " + ((val1 + val2) + val3)); System.out.println("sum 2 : " + (val1 + (val2 + val3))); System.out.println("sum 3 : " + (val1 + val2 + val3)); } } java Absorbtion sum 1 : sum 2 : sum 3 : Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 43

44 Comparing Floats: Idea Due to such problems, it is totally forbidden to compare oating point values using the == operator. This solution seen in many books fails most of the time: final float EPSILON = 10e-6;... if Math.abs(value1 - value2) < EPSILON { }... Problem: The value of the constant EPSILON depends on the magnitude of the tested values. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 44

45 Comparing Floats: Bitwise I One can try to compare both values bit-by-bit. For example: 1. Transform the oat values bitwise into integers (e.g. in Java using the method floattointbits()) 2. Mask the 2 LSB of each value 3. Compare the result. Question: Why does this comparison fail? Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 45

46 Comparing Floats: Bitwise II #include <stdio.h> typedef union { int intval; float floatval; } intfloat; int main() { intfloat testval; testval.intval = 0x3fffffff; printf("v1 as int : %x\n", testval.intval); printf("v1 as float : %f\n", testval.floatval); testval.intval = 0x ; printf("v2 as int : %x\n", testval.intval); printf("v2 as float : %f\n", testval.floatval); return 0; } Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 46

47 Comparing Floats: Bitwise III./a.out v1 as int : 3fffffff v1 as float : v2 as int : v2 as float : Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 47

48 Comparing Floats: Absolute Error I The main idea is to have an error margin which depends on the magnitude of the two values being compared. The simplest formula for that is: a b b < ε for a given ε value independent of the magnitude of a and b. If b is zero (or very near to zero) this direct computation fails! Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 48

49 Comparing Floats: Absolute Error II di a b if a is strictly equal to b then return true {for and NaN} else if a = 0 or b = 0 or ( a + b < min normal value) then {for very small value of a and b} return di < (ε min normal value) else return (di / min( a + b, max oat value)) < ε; end if Note: min normal value is the smallest value which can be stored without using a denormalized value. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 49

50 Total ordering predicate The standard provides a predicate totalorder which denes a total ordering for all oating point numbers in each format. The predicate agrees with the normal comparison operations when they say one oating point number is less than another. The normal predicate compares -0 and +0 as equal. The normal comparison operations however treat NaNs as unordered. Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 50

51 Conclusion Encoding Numbers Little and big endian Least and most signicant bytes (MSB/LSB) Real numbers Are composed of mantissa and exponent Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 51

52 Bibliography This course corresponds to chapter 5 and 6 of the course book: Assembly Language Step by Step (3rd Edition) Course of CPVR specialization: Introduction to Computer perception and virtual reality, Claude Fuhrer BFH-TI (FRC1) Bern University of Applied Sciences Berner Fachhochschule Haute école spécialisée bernoise 52

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